AIME(美国数学邀请赛)(2000-2018)答案
美国数学邀请赛(AIME Ⅱ 2016)试题与解答
美国数学邀请赛(AIME Ⅱ 2016)试题与解答殷琦涛;朱兆和【期刊名称】《上海中学数学》【年(卷),期】2016(000)005【总页数】5页(P44-48)【作者】殷琦涛;朱兆和【作者单位】200001 上海市格致中学;200001 上海市格致中学【正文语种】中文1.InitiallyAlex,Betty,andCharliehadatotalof444peanuts.Charliehadthemostp eanuts,andAlexhadtheleast.Thethreenumbersofpeanutsthateachpersonhad formedageometricprogression.Alexeats5ofhispeanuts,Bettyeats9ofherpean uts,andCharlieeats25ofhispeanuts.Nowthethreenumbersofpeanutseachpers onhasformsanarithmeticprogression.FindthenumberofpeanutsAlexhadinitia lly.译：亚历克斯、贝蒂和查利共有444颗花生，其中查利的花生最多，亚历克斯的花生最少，这三个人的花生数构成一个等比数列.亚历克斯吃掉5颗花生，贝蒂吃掉9颗花生，查利吃掉25颗花生，这时三个人的花生数构成一个等差数列.问：亚历克斯开始时有几颗花生？解：设亚历克斯、贝蒂、查利开始时各有a、aq、aq2(a∈N*,q>1)颗花生，根据题意可以得，即可以得，解得或(舍)，代入(1)得a=108.所以亚利克斯开始时有108颗花生.2.There is a 40% chance of rain on Saturday and a 30% chance of rain on Sunday. However, it is twice as likely to rain on Sunday if it rains on Saturday than if it does not rain on Saturday. The probability that it rains at least one day this weekend is, whereaandbare relatively prime positive integers. Finda+b.译：已知星期六下雨的概率是40%，星期天下雨的概率是30%.当星期六下雨时，星期天也下雨的概率是当星期六不下雨时、星期天下雨的概率的两倍.设周末至少有一天下雨的概率为，其中a,b是互质的正整数，求a+b的值.解：设当星期六不下雨时，星期天下雨的概率为p，则星期六下雨时，星期天也下雨的概率为2P.则有(1-40%)·p+40%·2p=30%，解得.周末两天都不下雨的概率为，所以周末两天至少有一天下雨的概率为.则a+b=107.3.Letx,yand z be real numbers satisfying the system.,,..译：设实数x,y,z满足,,.求的值.解：由题得，三式相加得3xyz+log5xyz=378.因为函数f(t)=3t+log5t是(0,+∞)上的单调递增函数，又f(125)=378，所以xyz=125.代入求得log5x=-90，log5y=-41，log5z=134.得.4.Ana×b×crectangular box is built froma·b·cunit cubes. Each unit cube is colored red, green, or yellow. Each of thealayers of size 1×b×cparallel to the (b×c)-faces of the box contains exactly 9 red cubes, exactly 12 green cubes, and some yellow cubes. Each of theblayers of sizea×1×cparallel to the (a×c)-faces of the box contains exactly 20 green cubes, exactly 25 yellow cubes, and some red cubes. Find the smallest possible volume ofthe box.译：一个a×b×c的盒子由a·b·c个单位正方体组成，每个单位正方体被染成红色、绿色或黄色.平行于盒子面的a层(大小为1×b×c)中，每一层刚好有9个红色单位正方体，12个绿色单位正方体和若干个黄色单位正方体.而平行于盒子面的b层(大小为a×1×c)中，每一层刚好有20个绿色单位正方体，25个黄色单位正方体和若干个红色单位正方体.求盒子体积的最小值.解：设平行于盒子b×c面的每一层中，黄色正方体有x个，平行于盒子a×c面的每一层中，红色正方体有y个.由题意，盒子中绿色正方体为12a或20b个，所以12a=20b，即b.同理,9a=by,ax=25b，所以有x=y=15.那么bc=21+x=36,ac=45+y=60.因a、b、c都是正整数，由知b≥3.V=abc=60b≥180，当b=3，c=12,a=5时等号成立.所以盒子体积的最小值为180.5.TriangleABC0has a right angle atC0. Its side lengths are pairwise relatively prime positive integers, and its perimeter isp. LetC1be the foot of the altitude to, and forn≥2, letCnbe the foot of the altitude toin △Cn-2Cn-1B. The sumCn-1Cn=6p. Findp.译：在△ABC0中，角C0是直角，三边长是两两互质的正整数，周长为p.C1是AB边上的高的垂足，对n≥2，设Cn是△Cn-2Cn-1B的边Cn-2B上高的垂足.已知，求p的值.解：设△ABC0的边长分别为a、b、c(a、b、c是两两互质的正整数)，且a2+b2=c2.如图1，设∠B=θ，则C0C1=asinθ，CnCn+1=Cn-1Cncosθ(n∈N*)，所以数列是以cosθ为公比的等比数列，首项C0C1=asinθ.，所以，所以ab=6(c-a)(a+b+c)，即7ab=6(c2-a2+bc)，又c2-a2=b2，所以7a=6(b+c)(1). 在(1)两边同乘(c-b)得7a(c-b)=6(b+c)(c-b)，即7a(c-b)=6a2，所以6a=7(c-b)(2).由(1)、(2)得a.设a=84k(k∈N*)，则a=84k,b=13k,c=85k，又a,b,c是两两互质的正整数，所以k=1，即a=84,b=13,c=85，所以p=182.6..Then, wheremandnare relatively prime positive integers. Findm+n.译：已知多项式，设，若，其中m,n是互质的正整数，求m+n的值.解(x2-2x+6)(x6-2x3+6)(x10-2x5+6)(x14-2x7+6)(x18-2x9+6),因为在5个小括号中x的偶次幂的系数为正数，奇数幂的系数为负数，在Q(x)的展开式中，x的偶次幂的项是由5个小括号中，在偶数个括号中取奇次幂，剩下的括号中取偶次幂相乘而得，所以展开式中x的偶次幂的系数为正，同理,展开式中x的奇次幂的系数为负.所以，即m+n=275.7.SquaresABCDandEFGHhave a common center and∥. The area ofABCDis 2016, and the area ofEFGHis a smaller positive integer. SquareIJKLis constructed so that each of its vertices lies on a side ofABCDand each vertex ofEFGHlies on a side ofIJKL. Find the difference between the largest and smallest positive integer values for the area ofIJKL.译：如图2，正方形ABCD与正方形EFGH有公共的中心，且AB∥EF.正方形ABCD的面积为2016，正方形EFGH的面积是比2016小的正整数.正方形IJKL的每个顶点在正方形ABCD的边上，且正方形EFGH的每个顶点在正方形IJKL的边上.求正方形IJKL面积的最大整数值与最小整数值的差.解：设AI=a,BI=b，则，当a=b时，.就在此时S正方形EFGH=504满足条件.依条件，正方形ABCD、正方形IJKL、正方形EFGH的面积成等比数列，设公比为，则，所以n2|2016=122×14，推得n|12.要使得S正方形IJKL最大，则m=n-1，所以，此式关于n递增，所以当n=12时，S正方形IJKL的最大值为1848，则所求差为1848-1008=840.8.Find the number of setsof three distinct positive integers with the property that the product ofa,bandcis equal to the product of 11, 21, 31, 41, 51 and 61.译：设集合是由三个不同的正整数组成的集合，满足a,b,c的积等于11、21、31、41、51、61的积，求满足条件的集合个数.解：abc=11×21×31×41×51×61=32×7×11×17×31×41×61.满足此式的有序数组(a,b,c)共有组.但{a,b,c}中的a、b、c无序且互不相等，所以数组(a,b,c)不等于(3,3,7×11×17×31×41×61)与(1,1,32×7×11×17×31×41×61)及其排列.故集合{a,b,c}的个数为.9.The sequences of positive integers 1,a2,a3,… and 1,b2,b3,… are an increasing arithmetic sequence and an increasing geometric sequence, respectively. Letcn=an+bn. There is an integerksuch thatck-1=100andck+1=1000. Findck.译：正整数数列1,a2,a3,…是递增的等差数列，正整数数列1,b2,b3,…是递增的等比数列，设cn=an+bn，若存在整数k，使得ck-1=100，且ck+1=1000，求ck 的值.解：设{an}的公差为d(d∈N*)，{bn}的公比为q(q∈N*,q≥2)，由题意得：，也即，依条件有k≥3.(1)k=3时，，解得q=9,d=90，此时ck=1+2d+q2=262；(2)k=4时，由(1)、(2)得⟹q4-2q2=801⟹(q2-1)2=802，此方程没有正整数解；(3)k≥5时，q5≤qk=999-kd<999，q只能取2或3，①当q=2时，，即(3k-8)d+603=0，又(3k-8)d>0，此式无解，②当q=3时，，即(8k-18)d+108=0，此式也无解.综上可知ck=262.10.TriangleABCis inscribed in circleω.Q. RaysCPandCQmeetωagain atSandT(other thanC), respectively.s. Findm+n.译：如图3，△ABC内接于圆ω，点P,Q在边AB上，且满足AP<AQ，射线CP,CQ分别交圆ω于点S,T(不同于点C).若AP=4,PQ=3,QB=6,BT=5，AS=7，那么，其中m,n是互质的正整数，求m+n的值.解：设∠ACP=α,∠PCQ=β,∠QCB=γ，△ACQ相似于△TBQ，所以，同理.⟹，⟹.又，所以，则，所以，即m+n=43.11.For positive integersNandk, defineNto bek-nice if there exists a positive integerasuch thatakhas exactlyNpositive divisors. Find the number of positive integers less than 1000 that are neither 7-nice nor 8-nice.译：对正整数N,k，若存在一个正整数a，使得ak的正因子个数刚好为N，则称N是k阶好数.在小于1000的正整数中，求既不是7阶好数，又不是8阶好数的数的个数.解：先证明引理,即N是k阶好数的充要条件是N≡1(modk).充分性：若N≡1(modk)，则N=km+1(k∈N)，设a=pm(p为素数)，则ak=pmk的正因子个数为km+1=N，所以N为k阶好数.必要性：设N是k阶好数，则存在a=p1α1·p2α2·…·pnαn，其中pi为两两互不相等的素数，αi∈N*(i=1,2,…n).则ak的正因子个数为，所以N≡1(modk).回到原题：在1，…，999中，7阶好数有个，8阶好数有个，56阶好数有个.所以满足条件的数有999-(143+125-18)=749个.12.The figure below shows a ring made of six small sections which you are to paint on a wall. You have four paint colors available and you will paint each of the six sections a solid color. Find the number of ways you can choose to paint the sections if no two adjacent sections can be painted with the same color.译：如图4，一个分成六个部分的环，用四种颜色染色，每个部分染一种颜色，且相邻的部分不能染同一种颜色，求不同的染色方法的种数.解：一个分成n个部分的环用四种颜色染色，相邻部分不同色，记不同的染色方法有an种，则a1=4,a2=4×3=12,a3=4×3×2=24.当n≥4时，将环的各个部分依次编号为1,2,3,…,n.在满足条件的染色方法下，先将第n部分去掉：(1)若第1部分与第n-1部分异色，则得到了一个有n-1部分的环的染色方法；反之，一个有n-1部分的环，在其第1部分与第n-1部分之间插入一段区域作为第n部分，则该部分有2种颜色可选.所以在这种情况下有2an-1种染色方法. (2)若第1部分与第n-1部分同色，则去掉第n部分后得到一个有n-2个部分的环的染色方法；反之，一个有n-2个部分的环，在其第n-2部分中插入一段区域，将第n-2部分分成两段，则新插入部分有三种不同的染色方法，所以在这种情况下有3an-2种染色方法.所以an=2an-1+3an-2(n≥4)，由a2=12,a3=24得a4=84,a5=240,a6=732.即不同的染色方法有732种.13.Beatrix is going to place six rooks on a 6×6 chessboard where both therows and columns are labeled 1 to 6; the rooks are placed so that no two rooks are in the same row or the same column. The value of a square is the sum of its row number and column number. The score of an arrangementof rooks is the least value of any occupied square. The average score over all valid configurations is, wherepandqare relatively prime positive integers. Findp+q.译：比阿特丽克斯在一个6×6的棋盘上放六个棋子，这些棋子中的任何两个既不同行也不同列，把棋盘的行和列用1，…,6编号.棋盘中每一格的“值”是它所在的行与列的编号数之和.在每种满足条件的放法中，每个棋子所占格子的“值”的最小值定义为这种放法的分数.若所有放法的分数的平均值是，其中p,q是互质的正整数，求p+q的值.解：用(a,b)表示棋子所占格子的位置，其中a、b分别表示该格所在的行、列的编号数.在每种满足条件的放法中，六个棋子的位置分别为(i,bi)(i=1,2,3,4,5,6)，其中b1,…,b6是1,…,6的一个排列.因为六个格子的“值”的平均值为=7，所以满足条件的放法的分数可能值为2,3,4,5,6,7.(1)分数为2时，b1=1，满足条件的放法有种.(2)分数为3时：①b1=2时，有种；②b1=3,4,5,6且b2=1时，有种，所以分数为3时的放法有120+96=216种.(3)分数为4时：①b1=3,则b2≥2，此时有种放法；②b1≥4时，又可分为b2=2，有放法，或b2≥3,b3=1，有种放法.所以分数为4的放法共有96+72+54=222种.(4)分数为5时：①b1=4时，则b2≥3且b3≥2，有种放法；②b1≥5，又可分为：b2=3,b3≥2，有种放法，或b2≥4,b3=2，有种放法，或b2≥4,b3≥3,b4=1，有种放法.所以分数为5的放法共有54+36+24+16=130种.(5)分数为6时：①b1=5时，则b2≥4,b3≥3,b4≥2，有种放法；②b1=6又可分为：b2=4,b3≥3,b4≥2，有种放法，或b2=5,b3=3,b4≥2，有种放法，或b2=5,b3=4,b4=2，有种放法，或b2=5,b3=4,b4=3,b5=1，只有1种放法.所以分数为6时的放法共有16+8+4+2+1=31种.(6)分数为7的放法仅有1种.综上=，所以p+q=371.14.Equilateral △ABChas side length 600. PointsPandQlie outside the plane of △ABCand are on opposite sides of the plane. Furthermore,PA=PB=PC, andQA=QB=QC, and the planes of △PABand △QABform a 120° dihedral angle (the angle between the two planes). There is a pointOwhose distance from each ofA,B,C,PandQisd. Findd.译：如图5，等边△ABC边长为600，P,Q两点位于△ABC所在平面异侧，满足PA=PB=PC，QA=QB=QC，二面角P-AB-Q为120°.若存在一点O，点O到A,B,C,P,Q的距离均为d，求d的值.解：设D为△ABC的中心，依条件P,D,Q三点共线，且O在PQ上.设PD=x,QD=y，可设x≥y，设AB中点为E，则，即100(x+y)=xy-3×104(1).又OA=OP，所以⟹xy=12×104，代入(1)得x+y=900，则.15.s. Findm+n.译：对1≤i≤215，设，且.正实数x1,x2,…,x216满足，且.设x2的最大值为，其中m,n是互质的正整数，求m+n的值.解，即,由柯西不等式，所以，由(1)，不等式取等号，所以，则，所以m+n=863.。
中国奥数最疯狂？强国靠的都是集训制度
中国奥数最疯狂？强国靠的都是集训制度奥数是一项国际赛事，可中国奥数的名声一直不好，非但不是与国际接轨，反而是中国教育的一个原罪。
有篇文章《对比中美日三国奥数的热与冷》里总结说，中国奥数是最疯狂最疲惫，日本奥数是很纯洁很低调，美国奥数是靠兴趣拼综合。
果然在某些人眼里，中国货永远像是“后妈生的”。
龙哥是上海人，分别当了多年中美奥数总教头的两位教授：熊斌和冯祖鸣，也都是上海人。
熊斌毕业于华东师大数学系，冯祖鸣的父母同样毕业于华东师大数学系，两人私下还是朋友。
龙哥自己从小就做熊老师出的题，虽然没参加奥数，但深感对学好数学大有裨益，后来大学也读了数学系。
冯祖鸣（图）自2003年起任美国国家队总教练，2014年，另一位华裔罗博深正式接棒。
美国奥数这么强，许多人又惊呼第一次听说美国人也玩这个。
熊斌教授早说了，中美两国的奥数水平差距不大，美国一直是奥数强国，1994年那一届所有6名选手全部获满分，这也是奥数史上唯一的一次。
“排名有先后，但实质并无差别，谁拿第一都很正常”。
当熊老师说无差别的时候，他说的可是体制问题哦。
几个奥数大国的竞赛体制颇为相似，中国、美国、俄罗斯、朝鲜、韩国、日本、越南……都有类似的集训制度，“完整的选拔、训练体制，以及国家的投入是奥数良好成绩的保证。
有些国家缺乏这样的体制，教练的投入也不够，就很难获得好成绩”。
原来强国都是集训制度这同一个妈生的嘛。
那说好的中国最疯狂呢？美国的奥数比赛比中国多，龙哥不是唬你。
想当年，龙哥高中毕业那会，有位复旦附中的姐姐，去了普林斯顿，很出名。
根据她给附中写的信中自述，她参加过美国数学才能测试（这是和AIME美国数学邀请赛齐名的美国数学国家队的选拔赛之一），H arvard-MIT数学邀请赛，据说她还以全美女生第一名的成绩入选美国数学冬令营。
龙哥写稿前又问了她美国奥数的情况，她说：“总而言之，层层选拔，层层培训，和中国一样。
”想进国家队？一般的流程是，先要参加全美数学竞赛（AMC）。
美国数学竞赛
竞赛题目的难度分布及运算量的不同
•我国数学竞赛题目缺乏应用性和技巧性，运算量较大，并且 一个题目的内容比较单一．美国数学竞赛题目应用性很强， 技巧性也较高，运算量比较小，同时能够够将多个知识点融 合到一个题目当中． •竞赛时间的不同 •我国中学数学竞赛由于题目较难，运算量较大，导致考试时 间较长．而美国题目难度稍低，运算量不大，同时有的竞赛 允许使用计算器，这样就减少了很多的竞赛时间．
IMO选拔
• • • • • 初中数学联赛 高中数学联赛 AMC8:11月中举行。 AMC10:2月初 AMC12:2月中。 美国数学邀请赛:3月底。 美国数学奥林匹克：4月的最后一周。 数学奥林匹克夏令营（MOSP） 国际数学奥林匹克（IMO)：7月
AMC通过其完善的测试体系一直承担着为美国培育世界数 学奥林匹克IMO选手的重任。AMC的研究人员透过AMC8、AMC 10、AMC12、AIME等一系列测验,找出绩优生参加USJMO和 USAMO，再从全美数十州筛选出12位精英，成立美国数学奥 林匹克夏令营（MOSP），最终从中选出美国参加IMO的6名选 手，组成IMO美国国家队。
美国中学数学竞赛的特点
• 1）组织者广泛、参赛者自愿 • 2）竞赛目的多样化 • 3）普及与选拔两项功能有机结合
• 4）竞赛形式多样化,评分规则多元化
中美数学竞赛的比较
•竞赛考察知识点的不同 •首先,美国考察的内容比我国考察的内容更接近实际生活。 •其次,由于美国高中阶段就已经讲解了我们在大学阶段学习 的内容，比如微积分、极限、级数等等，而我国高中只简单 的涉及一点微积分的知识，故在一些中学数学竞赛中，美国 考察的知识范围是大于同等级别中国数学竞赛考察知识的范 围的.美国中学数学教学的特点是广而浅，我国则是窄而 深．
2018美国数学竞赛(十、十二年级)
２０８年第２期３２０１８美国数学竞赛（十、十二年级）中图分类号Ｇ４２４７９文献标识码Ａ文章编号００６４２０２０００＋年级的扣除分每段视频均以０分开始计算段时间后此人看到他上传的视频得了９０１计算分其中投票的人有６５％喜欢此视频则有＝Ａ｜Ｂ＾（〇｜Ｄ｜｜Ｅ＾（Ａ）２００（Ｂ）３００（Ｃ４００（Ｄ５００（Ｅ６００２已知甲的苏打水容量比乙的多５０％有个整数使１＾值丙的苏打水容量比乙的多２５％则甲的苏打７ｎ４００的）水与丙的苏打水在容量上的关系为甲比丙多为整数（）Ａ３Ｂ）４Ｃ６Ｄ８Ｅ）９（Ｄ）７５％（Ｅ００％秒枚其中〇分硬币比５分硬币多三枚若硬３个单位血液的有效期为０亚币总价值为３２０分则此人的２５分硬币比５辛在月１日中午捐献了个单位血液则分硬币多枚他捐血液的有效期截止到（Ａ）０（Ｂ）Ｃ２（Ｄ）３（Ｅ）４图月２如图在等腰△中Ａ日Ｂ月２日９仙Ｃ狀＝４ＣＣ月２２日Ｄ）２月日中所有三角形均相似其中最小的三角形面（Ｅ）２月２日积为Ａ狀Ｃ的面积为４０则梯形洲以的４有代数几何数论三门数学课程若面积为（任意两门课程不能连排则名学生在天的６节课中有）种不同的排课方法（Ａ）３（Ｂ）６（Ｃ）２（Ｄ）１８Ｅ２４５甲乙丙三人结伴徒步旅行他们都想知道距最近的城镇有多远甲说，至少６英里”乙说“至多５英里”丙说“至多４英里”事实上三个人都说错了令ｄ表示他们与最近的城镇的距离则ｄ属于区间（）（Ａ）（０４（Ｂ）４５）（〇４６）Ｄ５６Ｅ５＋〇〇６某人将段视频上传到网站观看的人可以在网站上评价喜欢的加分不喜欢１０已知／４９ｘ＼／２５ｘ＝３＊＾Ｒ则７４９＊＋７２５＊的值为（（Ａ）８ＢＶ＾３＋８Ｃ９（Ｄ）２（Ｅ２＾＋４３２１１将七个标准的色子抛出后正面的数字之和是〇的概率为＃ｅｚ则的值为）（Ａ）４２Ｂ）４９（Ｃ）５６（Ｄ）６３（Ｅ）８４％＋３ｙ＝方程组３共有１２＾ｙ组解Ａｌ（Ｂ）２（Ｃ）３（Ｄ）４（Ｅ）８１３在ＲＡｌｉｆＣ中ｆｉＣ＝３ＡＣ＝４ＡＢ＝５将该三角形折叠使点４与Ｓ重合则折痕的长度为）Ａｌ＋ｆ（Ｂ）ｆｃ＋Ｄ＃（Ｅ）２ｑ０〇１４不大于ｆ的最大整数为）（Ａ）８０Ｂ）８（Ｃ）９６（Ｄ）９７（Ｅ）６２５１５如图２两个半径为５的等圆外切且均与个半径为３的大圆内切切点为记４５＝ｎｍｎＧＺ＋ｍｎ）＝则ｍ＋ｎ的值为Ａ２１Ｂ２９Ｃ５８Ｄ６９Ｅ９３１６在Ｒ△儿ＢＣ中Ｚ＝９０。
精选美国amc中文版试题及答案
2000到2012年A M C10美国数学竞赛P 0 A 0B 0 C0 D 0 全美中学数学分级能力测验(AMC 10)2000年 第01届 美国AMC10 (2000年2月 日 时间75分钟)1. 国际数学奥林匹亚将于2001年在美国举办，假设I 、M 、O 分别表示不同的正整数，且满足I ?M ?O =2001，则试问I ?M ?O 之最大值为 。
(A) 23 (B) 55 (C) 99 (D) 111 (E) 6712. 2000(20002000)为 。
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(A) 40 (B) 50 (C) 55 (D) 60 (E) 754. Candra 每月要付给网络公司固定的月租费及上网的拨接费，已知她12月的账单为12.48元，而她1月的账单为17.54元，若她1月的上网时间是12月的两倍，试问月租费是 元。
(A) 2.53 (B) 5.06 (C) 6.24 (D) 7.42 (E) 8.775. 如图M ，N 分别为PA 与PB 之中点，试问当P 在一条平行AB 的直在线移动时，下列各数值有 项会变动。
(a) MN 长 (b) △PAB 之周长 (c) △PAB 之面积 (d) ABNM 之面积 (A) 0项 (B) 1项 (C) 2项 (D) 3项 (E) 4项6. 费氏数列是以两个1开始，接下来各项均为前两项之和，试问在费氏数列各项的个位数字中， 最后出现的阿拉伯数字为 。
(A) 0 (B) 4 (C) 6 (D) 7 (E) 97. 如图，矩形ABCD 中，AD =1，P 在AB 上，且DP 与DB 三等分 ?ADC ，试问△BDP 之周长为 。
(完整版)(完整版)2018年(第59届)国际数学奥林匹克(IMO)竞赛试题及答案图片版
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AIME I 十年真题(附答案)
Problem 1Call a -digit number geometric if it has distinct digits which, when read from left to right, form a geometric sequence. Find the difference between the largest and smallest geometric numbers.Problem 2There is a complex number with imaginary part and a positive integer such thatFind .Problem 3A coin that comes up heads with probability and tails with probabilityindependently on each flip is flipped eight times. Suppose the probability of three heads and five tails is equal to of the probability of five heads and three tails. Let , where and are relatively prime positive integers. Find .Problem 4In parallelogram , point is on so that and pointis on so that . Let be the point of intersection of and . Find .Problem 5Triangle has and . Points and are located on and respectively so that , and is the angle bisector of angle . Let be the point of intersection of and , and let be the point on line for which is the midpoint of . If , find .Problem 6How many positive integers less than are there such that the equation has a solution for ? (The notation denotes the greatest integerthat is less than or equal to .)Problem 7The sequence satisfies and for . Let be the least integer greater than for which is an integer. Find . Problem 8Let . Consider all possible positive differences of pairsof elements of . Let be the sum of all of these differences. Find the remainder when is divided by .Problem 9A game show offers a contestant three prizes A,B and C, each of which is worth a whole number of dollars from $to $inclusive. The contestant wins the prizes by correctly guessing the price of each prize in the order A, B, C. As a hint, the digits of the three prices are given. On a particular day, the digits given were . Find the total number of possible guesses for all three prizes consistent with the hint.Problem 10The Annual Interplanetary Mathematics Examination (AIME) is written by a committee of five Martians, five Venusians, and five Earthlings. At meetings, committee members sit at a round table with chairs numbered from to in clockwise order. Committee rules state that a Martian must occupy chair and an Earthling must occupy chair , Furthermore, no Earthling can sit immediately to the left of a Martian, no Martian can sit immediately to the left of a Venusian, and no Venusian can sit immediately to the left of an Earthling.The number of possible seating arrangements for the committee is . Find .Problem 11Consider the set of all triangles where is the origin and and aredistinct points in the plane with nonnegative integer coordinates suchthat . Find the number of such distinct triangles whose area is a positive integer.Problem 12In right with hypotenuse , , , and is the altitude to . Let be the circle having as a diameter. Let be a point outside such that and are both tangent to circle . The ratio ofthe perimeter of to the length can be expressed in the form , where and are relatively prime positive integers. Find .Problem 13The terms of the sequence defined by for are positive integers. Find the minimum possible value of .Problem 14For , define , where . If and , find the minimum possible value for .Problem 15In triangle , , , and . Let be a point in the interior of . Let and denote the incenters of triangles and , respectively. The circumcircles of triangles and meet at distinct points and . The maximum possible area of can be expressed in the form , where , , and are positive integers and is not divisible by the square of any prime. Find .Of the students attending a school party, of the students are girls, and of the students like to dance. After these students are joined by more boy students, all of whom like to dance, the party is now girls. How many students now at the party like to dance?SolutionProblem 2Square has sides of length units. Isosceles triangle has base , and the area common to triangle and square is square units. Find the length of the altitude to in .SolutionProblem 3Ed and Sue bike at equal and constant rates. Similarly, they jog at equal and constant rates, and they swim at equal and constant rates. Ed covers kilometers after biking for hours, jogging for hours, and swimming for hours, while Sue covers kilometers after jogging for hours, swimming for hours, and biking for hours. Their biking, jogging, and swimming rates are all whole numbers of kilometers per hour. Find the sum of the squares of Ed's biking, jogging, and swimming rates.SolutionProblem 4There exist unique positive integers and that satisfy the equation. Find .SolutionProblem 5A right circular cone has base radius and height . The cone lies on its side on a flat table. As the cone rolls on the surface of the table without slipping, the point where the cone's base meets the table traces a circular arc centered at the point where the vertex touches the table. The cone first returns to its original position on the table after making complete rotations. The value of can be written in the form , where and are positive integers and is not divisible by the square of any prime. Find .A triangular array of numbers has a first row consisting of the odd integersin increasing order. Each row below the first has one fewer entry than the row above it, and the bottom row has a single entry. Each entry in any row after the top row equals the sum of the two entries diagonally above it in the row immediately above it. How many entries in the array are multiples ofProblem 7Let be the set of all integers such that . For example,is the set . How many of the setsdo not contain a perfect square?Problem 8Find the positive integer such thatProblem 9Ten identical crates each of dimensions ft ft ft. The first crate is placed flat on the floor. Each of the remaining nine crates is placed, in turn, flat on top of the previous crate, and the orientation of each crate is chosen at random.Let be the probability that the stack of crates is exactly ft tall, whereand are relatively prime positive integers. Find .Problem 10Let be an isosceles trapezoid with whose angle at thelonger base is . The diagonals have length , and point is at distances and from vertices and , respectively. Let be the foot of the altitude from to . The distance can be expressed in theform , where and are positive integers and is not divisible by the square of any prime. Find .Problem 11Consider sequences that consist entirely of 's and 's and that have the property that every run of consecutive 's has even length, and every run of consecutive 's has odd length. Examples of such sequences are , , and , while is not such a sequence. How many such sequences have length 14?Problem 12On a long straight stretch of one-way single-lane highway, cars all travel at the same speed and all obey the safety rule: the distance form the back of the car ahead to the front of the car behind is exactly one car length for each 15 kilometers per hour of speed or fraction thereof (Thus the front of a car traveling 52 kilometers per hour will be four car lengths behind the back of the car in front of it.) A photoelectric eye by the side of the road counts the number of cars that pass in one hour. Assuming that each car is 4 meters long and that the cars can travel at any speed, let be the maximum whole number of cars that can pass the photoelectric eye in one hour. Find the quotient when is divided by 10.Problem 13Let.Suppose that.There is a point for which for all such polynomials, where, , and are positive integers, and are relatively prime, and . Find .SolutionProblem 14Let be a diameter of circle . Extend through to . Point lies on so that line is tangent to . Point is the foot of the perpendicular from to line . Suppose , and let denote the maximum possible length of segment . Find .SolutionProblem 15A square piece of paper has sides of length . From each corner a wedge is cut in the following manner: at each corner, the two cuts for the wedge each start at distance from the corner, and they meet on the diagonal at an angle of (see the figure below). The paper is then folded up along the lines joining the vertices of adjacent cuts. When the two edges of a cut meet, they are taped together. The result is a paper tray whose sides are not at right angles to the base. The height of the tray, that is, the perpendicular distance between the plane of the base and the plane formed by the upper edges, can be written in the form , where and are positive integers, , and is not divisible by the th power of any prime. Find .How many positive perfect squares less than are multiples of 24?Problem 2A 100 foot long moving walkway moves at a constant rate of 6 feet per second. Al steps onto the start of the walkway and stands. Bob steps onto the start of the walkway two seconds later and strolls forward along the walkway at a constant rate of 4 feet per second. Two seconds after that, Cy reaches the start of the walkway and walks briskly forward beside the walkway at a constant rate of 8 feet per second. At a certain time, one of these three persons is exactly halfway between the other two. At that time, find the distance in feet between the start of the walkway and the middle person.Problem 3The complex number is equal to , where is a positive real number and . Given that the imaginary parts of and are the same, what is equal to?Problem 4Three planets orbit a star circularly in the same plane. Each moves in the same direction and moves at constant speed. Their periods are , , and . The three planets and the star are currently collinear. What is the fewest number of years from now that they will all be collinear again?Problem 5The formula for converting a Fahrenheit temperature to the correspondingCelsius temperature is An integer Fahrenheit temperature is converted to Celsius, rounded to the nearest integer, converted back to Fahrenheit, and again rounded to the nearest integer.For how many integer Fahrenheit temperatures between 32 and 1000 inclusive does the original temperature equal the final temperature?A frog is placed at the origin on the number line, and moves according to the following rule: in a given move, the frog advances to either the closest point with a greater integer coordinate that is a multiple of 3, or to the closest point with a greater integer coordinate that is a multiple of 13. A move sequence is a sequence of coordinates which correspond to valid moves, beginning with 0, and ending with 39. For example, is a move sequence. How many move sequences are possible for the frog?Problem 7LetFind the remainder when is divided by 1000. (is the greatest integer less than or equal to , and is the least integer greater than or equal to .)Problem 8The polynomial is cubic. What is the largest value of for which the polynomials andare both factors of ?Problem 9In right triangle with right angle , and . Its legsand are extended beyond and . Points and lie in the exterior of the triangle and are the centers of two circles with equal radii. The circle with center is tangent to the hypotenuse and to the extension of leg , the circle with center is tangent to the hypotenuse and to the extension of leg , and the circles are externally tangent to each other. The length of theradius of either circle can be expressed as , where and are relatively prime positive integers. Find .Problem 10In a 6 x 4 grid (6 rows, 4 columns), 12 of the 24 squares are to be shaded so that there are two shaded squares in each row and three shaded squares in each column. Let be the number of shadings with this property. Find the remainder when is divided by 1000.Problem 11For each positive integer , let denote the unique positive integer suchthat . For example, and . If find the remainder when is divided by 1000.Problem 12In isosceles triangle , is located at the origin and is located at (20,0). Point is in the first quadrant with and angle . If triangle is rotated counterclockwise about point until the image oflies on the positive -axis, the area of the region common to the original and the rotated triangle is in the form , where are integers. Find .Problem 13A square pyramid with base and vertex has eight edges of length 4.A plane passes through the midpoints of , , and . The plane's intersection with the pyramid has an area that can be expressed as . Find .Problem 14A sequence is defined over non-negative integral indexes in the following way:, .Find the greatest integer that does not exceedProblem 15Let be an equilateral triangle, and let and be points on sidesand , respectively, with and . Point lies on sidesuch that angle . The area of triangle is . The two possible values of the length of side are , where and are rational, and is an integer not divisible by the square of a prime. Find .In quadrilateral is a right angle, diagonal is perpendicular toand Find the perimeter ofProblem 2Let set be a 90-element subset of and let be the sum of the elements of Find the number of possible values ofProblem 3Find the least positive integer such that when its leftmost digit is deleted, the resulting integer is of the original integer.Problem 4Let be the number of consecutive 0's at the right end of the decimal representation of the product Find the remainder when is divided by 1000.Problem 5The number can be written aswhere and are positive integers. FindProblem 6Let be the set of real numbers that can be represented as repeating decimals of the form where are distinct digits. Find the sum of the elements ofAn angle is drawn on a set of equally spaced parallel lines as shown. The ratio of the area of shaded region to the area of shaded region is 11/5. Find the ratio of shaded region to the area of shaded regionProblem 8Hexagon is divided into five rhombuses, and as shown. Rhombuses and are congruent, and each has areaLet be the area of rhombus Given that is a positive integer, find the number of possible values forProblem 9The sequence is geometric with and common ratio where and are positive integers. Given thatfind the number of possible ordered pairsProblem 10Eight circles of diameter 1 are packed in the first quadrant of the coordinte plane as shown. Let region be the union of the eight circular regions. Line with slope 3, divides into two regions of equal area. Line 's equation can beexpressed in the form where and are positive integers whose greatest common divisor is 1. FindProblem 11A collection of 8 cubes consists of one cube with edge-length for each integer A tower is to be built using all 8 cubes according to the rules:Any cube may be the bottom cube in the tower.The cube immediately on top of a cube withedge-length must have edge-length at mostLet be the number of different towers than can be constructed. What is the remainder when is divided by 1000?Problem 12Find the sum of the values of such thatwhere is measured in degrees andProblem 13For each even positive integer let denote the greatest power of 2 that divides For example, and For each positive integerlet Find the greatest integer less than 1000 such that is a perfect square.Problem 14A tripod has three legs each of length 5 feet. When the tripod is set up, the angle between any pair of legs is equal to the angle between any other pair, and the top of the tripod is 4 feet from the ground In setting up the tripod, the lower 1 foot of one leg breaks off. Let be the height in feet of the top of the tripod from the ground when the broken tripod is set up. Then can be writtenin the form where and are positive integers and is not divisible by the square of any prime. Find (The notation denotes the greatest integer that is less than or equal to )Problem 15Given that a sequence satisfies and for all integers find the minimum possible value ofSix circles form a ring with with each circle externally tangent to two circles adjacent to it. All circles are internally tangent to a circle with radius 30. Let be the area of the region inside circle and outside of the six circles in thering. FindProblem 2For each positive integer let denote the increasing arithmetic sequence of integers whose first term is 1 and whose common difference is For example, is the sequence For how many values of does contain the term 2005?Problem 3How many positive integers have exactly three proper divisors, each of which is less than 50?Problem 4The director of a marching band wishes to place the members into a formation that includes all of them and has no unfilled positions. If they are arranged in a square formation, there are 5 members left over. The director realizes that if he arranges the group in a formation with 7 more rows than columns, there are no members left over. Find the maximum number of members this band can have.Problem 5Robert has 4 indistinguishable gold coins and 4 indistinguishable silver coins. Each coin has an engraving of one face on one side, but not on the other. He wants to stack the eight coins on a table into a single stack so that no two adjacent coins are face to face. Find the number of possible distunguishable arrangements of the 8 coins.Let be the product of the nonreal roots of FindProblem 7In quadrilateral andGiven that where and are positive integers, findProblem 8The equation has three real roots. Given that their sum is where and are relatively prime positive integers, findProblem 9Twenty seven unit cubes are painted orange on a set of four faces so that two non-painted faces share an edge. The 27 cubes are randomly arranged to form a cube. Given the probability of the entire surface area of thelarger cube is orange is where and are distinct primes and and are positive integers, findProblem 10Triangle lies in the Cartesian Plane and has an area of 70. The coordinates of and are and respectively, and the coordinates of are The line containing the median to side has slope Find the largest possible value ofProblem 11A semicircle with diameter is contained in a square whose sides have length8. Given the maximum value of is findFor positive integers let denote the number of positive integer divisors of including 1 and For example, and Define byLet denote the number of positive integers with odd, and let denote the number of positive integerswith even. FindProblem 13A particle moves in the Cartesian Plane according to the following rules:1. From any lattice point the particle mayonly move to or2. There are no right angle turns in the particle'spath.How many different paths can the particle take from to ?Problem 14Consider the points and There is a uniquesquare such that each of the four points is on a different side of Let be the area of Find the remainder when is divided by 1000.Problem 15Triangle has The incircle of the triangle evenly trisects the median If the area of the triangle is where and are integers and is not divisible by the square of a prime, findThe digits of a positive integer are four consecutive integers in decreasing order when read from left to right. What is the sum of the possible remainders when is divided by 37?Problem 2Set consists of consecutive integers whose sum is and set consists of consecutive integers whose sum is The absolute value of the difference between the greatest element of and the greatest element of is 99. FindProblem 3A convex polyhedron has 26 vertices, 60 edges, and 36 faces, 24 of which are triangular, and 12 of which are quadrilaterals. A space diagonal is a line segment connecting two non-adjacent vertices that do not belong to the same face. How many space diagonals does have?Problem 4A square has sides of length 2. Set is the set of all line segments that have length 2 and whose endpoints are on adjacent sides of the square. The midpoints of the line segments in set enclose a region whose area to the nearest hundredth is FindProblem 5Alpha and Beta both took part in a two-day problem-solving competition. At the end of the second day, each had attempted questions worth a total of 500 points. Alpha scored 160 points out of 300 points attempted on the first day, and scored 140 points out of 200 points attempted on the second day. Beta who did not attempt 300 points on the first day, had a positive integer score on each of the two days, and Beta's daily success rate (points scored divided by points attempted) on each day was less than Alpha's on that day. Alpha'stwo-day success ratio was 300/500 = 3/5. The largest possible two-daysuccess ratio that Beta could achieve is where and are relatively prime positive integers. What is ?Problem 6An integer is called snakelike if its decimal representationsatisfies if is odd and if is even. How many snakelike integers between 1000 and 9999 have four distinct digits?Problem 7Let be the coefficient of in the expansion of the productFindProblem 8Define a regular -pointed star to be the union of line segmentssuch thatthe points are coplanar and no threeof them are collinear,each of the line segments intersects at least oneof the other line segments at a point other than anendpoint,all of the angles at are congruent,all of the line segments arecongruent, andthe path turnscounterclockwise at an angle of less than 180degrees at each vertex.There are no regular 3-pointed, 4-pointed, or 6-pointed stars. All regular5-pointed stars are similar, but there are two non-similar regular 7-pointed stars. How many non-similar regular 1000-pointed stars are there?Problem 9Let be a triangle with sides 3, 4, and 5, and be a 6-by-7 rectangle. A segment is drawn to divide triangle into a triangle and a trapezoid and another segment is drawn to divide rectangle into a triangle and a trapezoid such that is similar to and is similar toThe minimum value of the area of can be written in the form where and are relatively prime positive integers. FindProblem 10A circle of radius 1 is randomly placed in a 15-by-36 rectangle so that the circle lies completely within the rectangle. Given that the probability thatthe circle will not touch diagonal is where and are relatively prime positive integers. FindProblem 11A solid in the shape of a right circular cone is 4 inches tall and its base has a 3-inch radius. The entire surface of the cone, including its base, is painted. A plane parallel to the base of the cone divides the cone into two solids, a smaller cone-shaped solid and a frustum-shaped solid in such a way that the ratio between the areas of the painted surfaces of and and the ratiobetween the volumes of and are both equal to Given thatwhere and are relatively prime positive integers, findProblem 12Let be the set of ordered pairs such that and and are both even. Given that the area of the graph of is where and are relatively prime positive integers, find The notation denotes the greatest integer that is less than or equal toThe polynomial has 34 complex roots of the form withand Given thatwhere and are relatively prime positive integers, findProblem 14A unicorn is tethered by a 20-foot silver rope to the base of a magician's cylindrical tower whose radius is 8 feet. The rope is attached to the tower at ground level and to the unicorn at a height of 4 feet. The unicorn has pulled the rope taut, the end of the rope is 4 feet from the nearest point on the tower, andthe length of the rope that is touching the tower is feet, where and are positive integers, and is prime. FindProblem 15For all positive integers , let and define a sequence as follows: and for all positive integers . Let be the smallest such that . (For example, and .) Let be the number of positive integers such that . Find the sum of the distinct prime factors of .SolutionGiven thatwhere and are positive integers and is as large as possible, findProblem 2One hundred concentric circles with radii are drawn in a plane. The interior of the circle of radius 1 is colored red, and each region bounded by consecutive circles is colored either red or green, with no two adjacent regions the same color. The ratio of the total area of the green regions to the area ofthe circle of radius 100 can be expressed as where and are relatively prime positive integers. FindProblem 3Let the set Susan makes a list as follows: for eachtwo-element subset of she writes on her list the greater of the set's two elements. Find the sum of the numbers on the list.Problem 4Given that and thatfindProblem 5Consider the set of points that are inside or within one unit of a rectangular parallelepiped (box) that measures 3 by 4 by 5 units. Given that the volume ofthis set is where and are positive integers, and andare relatively prime, findThe sum of the areas of all triangles whose vertices are also vertices of a 1 by 1 by 1 cube is where and are integers. FindProblem 7Point is on with and Point is not on so that and and are integers. Let be the sum of all possible perimeters of FindProblem 8In an increasing sequence of four positive integers, the first three terms form an arithmetic progression, the last three terms form a geometric progression, and the first and fourth terms differ by 30. Find the sum of the four terms.Problem 9An integer between 1000 and 9999, inclusive, is called balanced if the sum of its two leftmost digits equals the sum of its two rightmost digits. How many balanced integers are there?Problem 10Triangle is isosceles with and Point is in the interior of the triangle so that and Find the number of degrees inProblem 11An angle is chosen at random from the interval Let be the probability that the numbers and are not the lengths of the sides of a triangle. Given that where is the number of degrees in and and are positive integers with findProblem 12In convex quadrilateral andThe perimeter of is 640. Find (The notation means the greatest integer that is less than or equal to )Problem 13Let be the number of positive integers that are less than or equal to 2003 and whose base-2 representation has more 1's than 0's. Find the remainder when is divided by 1000.Problem 14The decimal representation of where and are relatively primepositive integers and contains the digits 2, 5, and 1 consecutively, and in that order. Find the smallest value of for which this is possible.Problem 15In and Let be the midpoint of and let be the point on such that bisects angle Let be the point on such that Suppose that meets at The ratio can be written in the form where and are relatively prime positive integers. FindMany states use a sequence of three letters followed by a sequence of three digits as their standard license-plate pattern. Given that each three-letter three-digit arrangement is equally likely, the probability that such a license plate will contain at least one palindrome (a three-letter arrangement or a three-digit arrangement that reads the same left-to-right as it does right-to-left)is , where and are relatively prime positive integers. Find .Problem 2The diagram shows twenty congruent circles arranged in three rows and enclosed in a rectangle. The circles are tangent to one another and to the sides of the rectangle as shown in the diagram. The ratio of the longer dimension of the rectangle to the shorter dimension can be written as, where and are positive integers. Find .Problem 3Jane is 25 years old. Dick is older than Jane. In years, where n is a positive integer, Dick's age and Jane's age will both be two-digit number and will have the property that Jane's age is obtained by interchanging the digits of Dick's age. Let be Dick's present age. How many ordered pairs of positive integersare possible?。
2002年美国数学邀请赛(AIME)
2002I 1Many states use a sequence of three letters followed by a sequence of three digits as their standard license-plate pattern.Given that each three-letter three-digit arrangement is equally likely,the probability that such a license plate will contain at least one palindrome (a three-letter arrangement or a three-digit arrangement that reads the same left-to-right as it does right-to-left)is m/n ,where m and n are relatively prime positive integers.Find m +n .2The diagram shows twenty congruent circles arranged in three rows and enclosed in a rect-angle.The circles are tangent to one another and to the sides of the rectangle as shown in the diagram.The ratio of the longer dimension of the rectangle to the shorter dimension can be written as 12(√p −q ),where p and q are positive integers.Find p +q.[img]6601[/img]3Jane is 25years old.Dick is older than Jane.In n years,where n is a positive integer,Dick’s age and Jane’s age will both be two-digit number and will have the property that Jane’s age is obtained by interchanging the digits of Dick’s age.Let d be Dick’s present age.How many ordered pairs of positive integers (d,n )are possible?4Consider the sequence defined by a k =1k 2+k for k ≥1.Given that a m +a m +1+···+a n −1=1/29,for positive integers m and n with m <n ,find m +n.5Let A 1,A 2,A 3,...,A 12be the vertices of a regular dodecagon.How many distinct squares in the plane of the dodecagon have at least two vertices in the set {A 1,A 2,A 3,...,A 12}?6The solutions to the system of equationslog 225x +log 64y =4log x 225−log y 64=1are (x 1,y 1)and (x 2,y 2).Find log 30(x 1y 1x 2y 2).7The Binomial Expansion is valid for exponents that are not integers.That is,for all real numbers x,y,and r with |x |>|y |,(x +y )r =x r +rx r −1+r (r −1)2x r −2y 2+r (r −1)(r −2)3!x r −3y 3+···What are the first three digits to the right of the decimal point in the decimal representation of 102002+1 10/7?/This file was downloaded from the AoPS −MathLinks Math Olympiad Resources Page Page 1http://www.mathlinks.ro/20028Find the smallest integer k for which the conditions (1)a 1,a 2,a 3,...is a nondecreasing sequence of positive integers (2)a n =a n −1+a n −2for all n >2(3)a 9=k are satisfied by more than one sequence.9Harold,Tanya,and Ulysses paint a very long picket fence.Harold starts with the first picket and paints every h th picket;Tanya starts with the second picket and paints everth t th picket;and Ulysses starts with the third picket and paints every u th picket.Call the positive integer 100h +10t +u paintable when the triple (h,t,u )of positive integers results in every picket being painted exaclty once.Find the sum of all the paintable integers.10In the diagram below,angle ABC is a right angle.Point D is on BC ,and AD bisects angleCAB .Points E and F are on AB and AC ,respectively,so that AE =3and AF =10.Given that EB =9and F C =27,find the integer closest to the area of quadrilateral DCF G.[img]6604[/img]11Let ABCD and BCF G be two faces of a cube with AB =12.A beam of light emanates fromvertex A and reflects offface BCF G at point P,which is 7units from BG and 5units from BC.The beam continues to be reflected offthe faces of the cube.The length of the light path from the time it leaves point A until it next reaches a vertex of the cube is given by m √n,where m and n are integers and n is not divisible by the square of any prime.Find m +n.12Let F (z )=z +i z −i for all complex numbers z =i,and let z n =F (z n −1)for all positive integers n.Given that z 0=1137+i and z 2002=a +bi,where a and b are real numbers,find a +b.13In triangle ABC the medians AD and CE have lengths 18and 27,respectively,and AB =24.Extend CE to intersect the circumcircle of ABC at F .The area of traingle AF B is m √n ,where m and n are positive integers and n is not divisible by the square of any prime.Find m +n .14A set S of distinct positive integers has the following property:for every integer x in S ,thearithmetic mean of the set of values obtained by deleting x from S is an integer.Given that 1belongs to S and that 2002is the largest element of S ,what is the greatet number of elements that S can have?15Polyhedron ABCDEF G has six faces.Face ABCD is a square with AB =12;face ABF Gis a trapezoid with AB parallel to GF ,BF =AG =8,and GF =6;and face CDE has CE =DE =14.The other three faces are ADEG,BCEF,and EF G.The distance from E to face ABCD is 12.Given that EG 2=p −q √r,where p,q,and r are positive integers and r is not divisible by the square of any prime,find p +q +r.2002II 1Given that(1)x and y are both integers between 100and 999,inclusive;(2)y is the number formed by reversing the digits of x;and(3)z =|x −y |.How many distinct values of z are possible?2Three vertices of a cube are P =(7,12,10),Q =(8,8,1),and R =(11,3,9).What is the surface area of the cube?3It is given that log 6a +log 6b +log 6c =6,where a,b,and c are positive integers that form an increasing geometric sequence and b −a is the square of an integer.Find a +b +c.4Patio blocks that are hexagons 1unit on a side are used to outline a garden by placing the blocks edge to edge with n on each side.The diagram indicates the path of blocks around the garden when n =5.[img]6613[/img]If n =202,then the area of the garden enclosed by the path,not including the path itself,is m (√3/2)square units,where m is a positive integer.Find the remainder when m is divided by 1000.5Find the sum of all positive integers a =2n 3m ,where n and m are non-negative integers,for which a 6is not a divisor of 6a .6Find the integer that is closest to 100010000n =31n −4.7It is known that,for all positive integers k,12+22+32+···+k 2=k (k +1)(2k +1)6.Find the smallest positive integer k such that 12+22+32+···+k 2is a multiple of 200.8Find the least positive integer k for which the equation 2002n =k has no integer solutions for n.(The notation x means the greatest integer less than or equal to x.)20029Let S be the set {1,2,3,...,10}.Let n be the number of sets of two non-empty disjoint subsets of S .(Disjoint sets are defined as sets that have no common elements.)Find the remainder obtained when n is divided by 1000.10While finding the sine of a certain angle,an absent-minded professor failed to notice that hiscalculator was not in the correct angular mode.He was lucky to get the right answer.The two least positive real values of x for which the sine of x degrees is the same as the sine of x radians are mπn −πand pπq +π,where m,n,p and q are positive integers.Find m +n +p +q.11Two distinct,real,infinite geometric series each have a sum of 1and have the same secondterm.The third term of one of the series is 1/8,and the second term of both series can bewritten in the form √m −n p ,where m,n,and p are positive integers and m is not divisible by the square of any prime.Find 100m +10n +p.12A basketball player has a constant probability of .4of making any given shot,independantof previous shots.Let a n be the ratio of shots made to shots attempted after n shots.The probability that a 10=.4and a n ≤.4for all n such that 1≤n ≤9is given to be p a q b r/(s c ),where p,q,r,and s are primes,and a,b,and c are positive integers.Find (p +q +r +s )(a +b +c ).13In triangle ABC,point D is on BC with CD =2and DB =5,point E is on AC withCE =1and EA =32,AB =8,and AD and BE intersect at P.Points Q and R lie on AB so that P Q is parallel to CA and P R is parallel to CB.It is given that the ratio of the area of triangle P QR to the area of triangle ABC is m/n,where m and n are relatively prime positive integers.Find m +n.14The perimeter of triangle AP M is 152,and the angle P AM is a right angle.A circle ofradius 19with center O on AP is drawn so that it is tangent to AM and P M.Given that OP =m/n,where m and n are relatively prime positive integers,find m +n.15Circles C 1and C 2intersect at two points,one of which is (9,6),and the product of the radiiis 68.The x-axis and the line y =mx ,where m >0,iare tangent to both circles.It is given that m can be written in the form a √b/c,where a,b,and c are positive integers,b is not divisible by the square of any prime,and a and c are relatively prime.Find a +b +c.。
2003美国数学邀请赛(AIME)试题
2003AIME1． 设有三个正整数的乘积N 是这三个正整数之和的6倍，且其中一个数是另外两个数的和，则N 的所有可能值之和是多少？2． 设N 表示每一位数上的数字都不同的正整数，且为8的最大整数倍数，则N 除以1000所得的余数是多少？3． 定义一个“好字”为只由三个字母A ，B ，C 所组成的序列（A ，B ，C 不必全部出现在序列中），其中A 不能紧接在B 之后，B 不能紧接在C 之后且C 不能紧接在A 之后。
试问7个字母的“好字”有多少个？4．在一个正四面体中，以其四个面的中心为顶点可形成一个较小的四面体，若小四面体和大四面体的体积之比为mn，其中m ，n 为互质的正整数，则m+n=？ 5．一圆柱形木头其直径为12英寸，利用两次贯穿此木头的平面切割，可得一个契形木块。
第一次切割垂直于此圆柱的中心线，第二次切割所在平面与第一次切割所在平面所夹的二面角为45º，且此二平面之交线恰与圆柱交于一点设所得之契形木块的体积为n π立方英寸，其中n 为正整数，求n 。
6．在ΔABC 中，AB=13，BC=14，AC=15。
且G 为三中线的交点，若以G 为中心，将ΔABC 旋转180º后，A ，B ，C 三点分别移动到A ’,B’,C’.那么ΔABC 与ΔA ’B ’C ’所盖住的平面面积是多少？7．已知菱形ABCD ，ΔABD 与ΔACD 的外接圆半径分别为12。
5与2。
5，试求此菱形面积。
8．由某两个等差数列之对应项相乘所得的数列为1440，1716，1848，…试求此数列的第8项。
9．设多项式6532()P x x x x x =---，且432()1Q x x x x =---，已知1234,,,z z z z 为()0Q x =的根，求1234()()()()P z P z P z P z +++=？10．两个正整数相差60，它们平方根的和是某个非完全平方整数的平方根，试问这样的两个正整数之和其最大可能值是多少？11．ΔABC 为一直角三角形。
2001年第19届美国数学邀请赛AIME
2001年第19届美国数学邀请赛(AIME) 1.令N为具有如下性质的最大正整数:从左至右读N时,每相邻两个数字所组成的二位数均为一完全平方数,试问N的最左边三个数字是什么?2.某高级中学共有2001个学生,其中每一位学生都要选修西班牙语或法语,有些学生同时选修这两种语言.已知选修西班牙语的学生占全校学生总人数的百分比介于80%与85%之间,而选修法语的则介于30%与40%之间.若令m及M分别表示可能选修这两种语言的最少学生人数与最多学生人数,则M-m=?3.已知x1=211,x2=375,x3=420,x4=523,且x n=x n-1-x n-2+x n-3+x n-4(n≥5),试求x531+x753+x975=?4.令R=(8,6),两点P及Q分别在直线8y=15x及10y=3x上,且点R为PQ的中点.若线段PQ之长为mn(m与n为互质的正整数),则m+n=?5.若一组正数所成集合中存在三个相异元素可构成三角形的三边,称此集合为具有“三角形性质”,设有连续正整数之集合{4,5,6,…,n},其所有含十个元素的子集合都具有三角形性质,则n的最大可能值为何?6.设AB CD为圆内接正方形,且正方形EFGH的顶点E及F在边CD上,顶点G及H在圆上.若S正方形EFGHS正方形AB CD=mn(m与n为互质的正整数,且m<n)时,试求10n+m.7.设△PQ R为直角三角形,PQ=90,PR=120,Q R=150.令△PQ R的内切圆为C1,作直线S T⊥PR,分别交PR及Q R于点S及T,且S T与圆C1相切.又作直线U V⊥PQ,分别交PQ及Q R于点U及V,且U V与圆C1相切.令△Q U V的内切圆为C2,而△RS T的内切圆为C3.当两圆C1与C2的圆心距离可以表示成10n时,则n =?8.设函数f具有如下性质:f(3x)=3f(x)对于所有的正实数值x 都成立,且f(x)=1-|x-2|(1≤x≤3),试求满足f(x)=f(2001)之最小值.9.设有一3×3单位正方形所成之方格(由9个边长为1的正方形所构成的大正方形).每一个单位正方形方格都要涂上蓝色或红色,两种颜色被使用之机会都相等.若3×3单位正方形所成方格中没有一个2×2单位正方形所成方格为红色的概率是mn(m 与n为互质的正整数),试求m+n=?10.1001的正整数倍数中,有多少个可以表示成10i-10j的形式(其中i,j表整数,且0≤i<j≤99).11.S队在某一足球联赛中要和其他六个队的每一个都要比赛一场,已知S队在六场比赛中任何一次比赛打胜、打败或打成平手的概率都是13,设S队在打完这六场比赛后,胜的次数多于败的次数的概率为mn(m 与n为互质的正整数),则m+n=?12.连接一三角形三边中点所得之三角44数学通讯 2003年第17期2002年第20届美国数学邀请赛(AIME)试题 1.已知1)x与y均为介于100与999之间(含100与999)的整数.2)y可由x的各位数字反向排列而得.3)z=|x-y|.试问z共有多少个不同的可能值?2.已知一正方体的三个顶点为P(7, 12,10),Q(8,8,1)与R(11,3,9).试问此正方体的表面积是多少?3.已知log6a+log6b+log6c=6,其中a,b,c为正整数.若a,b,c为等比递增数列,且b-a为一完全平方数,试求a+b +c.4.中南美洲国家房子庭院中要建造花第4题图园时,都会在花园的每边使用n个边长为1单位的正六边形砖块,边连边地铺设出花园的轮廓.右图所示为当n=5时围绕着花园的砖块小路.若n=202时,小路所围绕的花园面积(不含小路本身)为32m平方单位,其中m 表正整数.试求m除以1000所得之余数.5.设a=2n・3m,其中n与m皆表大于或等于0的整数.则满足a6不是6a的因数的所有正整数a的和是多少? 形,称为中点三角形.现定义一序列多面体Pi如下:令P0表示体积为1的正四面体,设Pi为以每一个面上的中点三角形为一面再向外作正四面体所构成的新多面体为Pi+1.若P3的体积为mn(m与n为互质的正整数),则m+n=?13.在四边形AB CD中,∠BA D=∠A DC,∠ABD=∠B CD,A C=8,BD=10,B C=6,若线段CD的长可以表示为mn (m与n为互质的正整数),则m+n=?14.设满足z28-z8-1=0及|z|=1的复数共有2n个,这些复数的三角形式为zm =cosθm+i sinθm(0≤θ1<θ2<…<θ2n< 360°),且θi是以度为单位的,试求:θ2+θ4 +…+θ2n=?15.设EFGH,EFDC与EHB C为正方体的三个相邻接的正方形面,其中EC=8,令A表示正方体的第八个顶点,令I,J及K 三点分别在EF,EH及EC上,且EI=EJ =E K=2.设从正方体钻出一个通道后所得之立体为S,此通道含有边IJ,J K及KI,且其壁面平行于A E,设立体连同通道的表面积为m+n p(m与n皆表正整数,且p不能被任何质数之平方所整除),试求m+n+ p=?试题答案1.816.2.298.3.898.4.067.5.253.6.251.7.725.8.429.9.929. 10.784. 11.341. 12.101.13.069. 14.840. 15.417.提供 齐世荫(武汉市第二中学 430010)542003年第17期 数学通讯。
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AMC_美国数学竞赛_2000_AMC_10__试题及答案解析
USA AMC 10 20001In the year , the United States will host the International Mathematical Olympiad. Let , , and be distinct positive integers such that the product . What's the largest possible value of the sum ?SolutionThe sum is the highest if two factors are the lowest.So, and .2Solution.3Each day, Jenny ate of the jellybeans that were in her jar at the beginning of the day. At the end of the second day, remained. How many jellybeans were in the jar originally?Solution4Chandra pays an online service provider a fixed monthly fee plus an hourly charge for connect time. Her December bill was , but in January her bill was because she used twice as much connect time as in December. What is the fixxed monthly fee?SolutionLet be the fixed fee, and be the amount she pays for the minutes she used in the first month.We want the fixed fee, which is5Points and are the midpoints of sides and of . As moves along a line that is parallel to side , how many of the four quantities listed below change?(a) the length of the segment(b) the perimeter of(c) the area of(d) the area of trapezoidSolution(a) Clearly does not change, and , so doesn't change either.(b) Obviously, the perimeter changes.(c) The area clearly doesn't change, as both the base and its corresponding height remain the same.(d) The bases and do not change, and neither does the height, so the trapezoid remains the same.Only quantity changes, so the correct answer is .6The Fibonacci Sequence starts with two 1s and each term afterwards is the sum of its predecessors. Which one of the ten digits is the last to appear in thet units position of a number in the Fibonacci Sequence?SolutionThe pattern of the units digits areIn order of appearance:.is the last.7In rectangle , , is on , and and trisect . What is the perimeter of ?Solution.Since is trisected, .Thus,.Adding, .8At Olympic High School, of the freshmen and of the sophomores took the AMC-10. Given that the number of freshmen and sophomore contestants was the same, which of the following must be true?There are five times as many sophomores as freshmen.There are twice as many sophomores as freshmen.There are as many freshmen as sophomores.There are twice as many freshmen as sophomores.There are five times as many freshmen as sophomores.SolutionLet be the number of freshman and be the number of sophomores.There are twice as many freshmen as sophomores.9If , where , thenSolution, so ...10The sides of a triangle with positive area have lengths , , and . The sides of a second triangle with positive area have lengths , , and . What is the smallest positive number that is not a possible value of ?SolutionFrom the triangle inequality, and . The smallest positive number not possible is , which is .11Two different prime numbers between and are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained?SolutionTwo prime numbers between and are both odd.Thus, we can discard the even choices.Both and are even, so one more than is a multiple of four.is the only possible choice.satisfy this, .12Figures , , , and consist of , , , and nonoverlapping unit squares, respectively. If the pattern were continued, how many nonoverlapping unit squares would there be in figure 100?SolutionSolution 1We have a recursion:.I.E. we add increasing multiples of each time we go up a figure. So, to go from Figure 0 to 100, we add.Solution 2We can divide up figure to get the sum of the sum of the first odd numbers and the sum of the first odd numbers. If you do not see this, here is the example for :The sum of the first odd numbers is , so for figure , there are unit squares. We plug in to get , which is choice13There are 5 yellow pegs, 4 red pegs, 3 green pegs, 2 blue pegs, and 1 orange peg to be placed on a triangular peg board. In how many ways can the pegs be placed so that no (horizontal) row or (vertical) column contains two pegs of the same color?SolutionIn each column there must be one yellow peg. In particular, in the rightmost column, there is only one peg spot, therefore a yellow peg must go there.In the second column from the right, there are two spaces for pegs. One of them is in the same row as the corner peg, so there is only one remaining choice left for the yellow peg in this column.By similar logic, we can fill in the yellow pegs as shown:After this we can proceed to fill in the whole pegboard, so there is only arrangement of the pegs. The answer is14Mrs. Walter gave an exam in a mathematics class of five students. She entered the scores in random order into a spreadsheet, which recalculated the class average after each score was entered. Mrs. Walter noticed that after each score was entered, the average was always an integer. The scores (listed in ascending order) were , , , , and . What was the last score Mrs. Walter entered?SolutionThe sum of the first scores must be even, so we must choose evens or the odds to be the first two scores.Let us look at the numbers in mod .If we choose the two odds, the next number must be a multiple of , of which there is none.Similarly, if we choose or , the next number must be a multiple of , of which there is none.So we choose first.The next number must be 1 in mod 3, of which only remains.The sum of the first three scores is . This is equivalent to in mod .Thus, we need to choose one number that is in mod . is the only one that works.Thus, is the last score entered.15Two non-zero real numbers, and , satisfy . Which of the following is a possible value of ?SolutionSubstituting , we get16The diagram shows lattice points, each one unit from its nearest neighbors. Segment meets segment at . Find the length of segment .SolutionSolution 1Let be the line containing and and let be the line containing and . If we set the bottom left point at , then , , , and .The line is given by the equation . The -intercept is , so . We are given two points on , hence we cancompute the slope, to be , so is the lineSimilarly, is given by . The slope in this case is , so . Plugging in the point gives us , so is the line .At , the intersection point, both of the equations must be true, soWe have the coordinates of and , so we can use the distance formula here:which is answer choiceSolution 2Draw the perpendiculars from and to , respectively. As it turns out, . Let be the point on for which ., and , so by AA similarity,By the Pythagorean Theorem, we have ,, and . Let , so , thenThis is answer choiceAlso, you could extend CD to the end of the box and create two similar triangles. Then use ratios and find that the distance is 5/9 of the diagonal AB. Thus, the answer is B.17Boris has an incredible coin changing machine. When he puts in a quarter, it returns five nickels; when he puts in a nickel, it returns five pennies; and when he puts in a penny, it returns five quarters. Boris starts with just one penny. Which of the following amounts could Boris have after using the machine repeatedly?SolutionConsider what happens each time he puts a coin in. If he puts in a quarter, he gets five nickels back, so the amount of money he has doesn't change. Similarly, if he puts a nickel in the machine, he gets five pennies back and the money value doesn't change. However, if he puts a penny in, he gets five quarters back, increasing the amount of money he has by cents.This implies that the only possible values, in cents, he can have are the ones one more than a multiple of . Of the choices given, the only one is18Charlyn walks completely around the boundary of a square whose sides are each km long. From any point on her path she can see exactly km horizontally in all directions. What is the area of the region consisting of all points Charlyn can see during her walk, expressed in square kilometers and rounded to the nearest whole number?SolutionThe area she sees looks at follows:The part inside the walk has area . The part outside the walk consists of four rectangles, and four arcs. Each of the rectangles has area . The four arcs together form a circle with radius . Therefore the total area she can see is, which rounded to the nearest integer is .19Through a point on the hypotenuse of a right triangle, lines are drawn parallel to the legs of the triangle so that the trangle is divided into a square and two smaller right triangles. The area of one of the two small right triangles is times the area of the square. The ratio of the area of the other small right triangle to the area of the square isSolutionLet the square have area , then it follows that the altitude of one of the triangles is . The area of the other triangle is .By similar triangles, we haveThis is choice(Note that this approach is enough to get the correct answer in the contest. However, if we wanted a completely correct solution, we should also note that scaling the given triangle times changes each of the areas times, and therefore it does not influence the ratio of any two areas. This is why we can pick the side of the square.)20Let , , and be nonnegative integers such that . What is the maximum value of ?SolutionThe trick is to realize that the sum is similar to the product .If we multiply , we get.We know that , therefore.Therefore the maximum value of is equal to the maximum value of . Now we will find this maximum.Suppose that some two of , , and differ by at least . Then this triple is surely not optimal.Proof: WLOG let . We can then increase the value ofby changing and .Therefore the maximum is achieved in the cases where is a rotation of . The value of in this case is . And thus the maximum of is.21If all alligators are ferocious creatures and some creepy crawlers are alligators, which statement(s) must be true?I. All alligators are creepy crawlers.II. Some ferocious creatures are creepy crawlers.III. Some alligators are not creepy crawlers.SolutionWe interpret the problem statement as a query about three abstract concepts denoted as "alligators", "creepy crawlers" and "ferocious creatures". In answering the question, we may NOT refer to reality -- for example to the fact that alligators do exist.To make more clear that we are not using anything outside the problem statement, let's rename the three concepts as , , and .We got the following information:▪If is an , then is an .▪There is some that is a and at the same time an .We CAN NOT conclude that the first statement is true. For example, the situation "Johnny and Freddy are s, but only Johnny is a "meets both conditions, but the first statement is false.We CAN conclude that the second statement is true. We know that there is some that is a and at the same time an . Pick one such and call it Bobby. Additionally, we know that if is an , then is an. Bobby is an , therefore Bobby is an . And this is enough to prove the second statement -- Bobby is an that is also a .We CAN NOT conclude that the third statement is true. For example, consider the situation when , and are equivalent (represent the same set of objects). In such case both conditions are satisfied, but the third statement is false.Therefore the answer is .22One morning each member of Angela's family drank an -ounce mixture of coffee with milk. The amounts of coffee and milk varied from cup to cup, but were never zero. Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee. How many people are in the family?SolutionThe exact value "8 ounces" is not important. We will only use the fact that each member of the family drank the same amount.Let be the total number of ounces of milk drank by the family and the total number of ounces of coffee. Thus the whole family drank a total of ounces of fluids.Let be the number of family members. Then each family member drank ounces of fluids.We know that Angela drank ounces of fluids.As Angela is a family member, we have .Multiply both sides by to get .If , we have .If , we have .Therefore the only remaining option is .23When the mean, median, and mode of the list are arranged in increasing order, they form a non-constant arithmetic progression. What is the sum of all possible real values of ?SolutionAs occurs three times and each of the three other values just once, regardless of what we choose the mode will always be .The sum of all numbers is , therefore the mean is .The six known values, in sorted order, are . From this sequence we conclude: If , the median will be . If , the median will be . Finally, if , the median will be .We will now examine each of these three cases separately.In the case , both the median and the mode are 2, therefore we can not get any non-constant arithmetic progression.In the case we have , because. Therefore our three values inorder are . We want this to be an arithmetic progression. From the first two terms the difference must be . Therefore thethird term must be .Solving we get the only solution for this case: . The case remains. Once again, we have ,therefore the order is . The only solution is when , i. e., .The sum of all solutions is therefore .24Let be a function for which . Find the sum of all values of for which .SolutionIn the definition of , let . We get: . Aswe have , we must have , in other words .One can now either explicitly compute the roots, or use Vieta's formulas. According to them, the sum of the roots of is . In our case this is .(Note that for the above approach to be completely correct, we should additionally verify that there actually are two distinct real roots. This is, for example, obvious from the facts that and .)25In year , the day of the year is a Tuesday. In year , the day is also a Tuesday. On what day of the week did the of year occur?SolutionClearly, identifying what of these years may/must/may not be a leap year will be key in solving the problem.Let be the day of year , the day of year and the day of year .If year is not a leap year, the day will bedays after . As , that would be a Monday.Therefore year must be a leap year. (Then is days after .) As there can not be two leap years after each other, is not a leap year. Therefore day is days after . We have . Therefore is weekdays before , i.e., is a.(Note that the situation described by the problem statement indeed occurs in our calendar. For example, for we have =Tuesday, October 26th 2004, =Tuesday, July 19th, 2005 and =Thursday, April 10th 2003.)。
2001AIME美国数学邀请赛
2001 AIME11、 求所有这样的两位数之和，它们满足能被其中任意数码整除.2、 不同实数的有限集S 满足如下性质：{}1S 的平均数比S 的平均数小13，且{}2001S 的平均数比S 的平均数大27，求S 的平均数.3、 找出方程200120011()02xx +-=的所有根，包括实根与虚根，假定方程没有复根.4、 在A B C ∆中，60,45A B ︒︒∠=∠=.A ∠的角平分线交B C 于点T ，且24A T =.A B C∆的面积可以写成a +其中a b c ，，都是正整数，且c 不能被任何素数的平方整除.求a b c ++.5、 椭圆2244x y +=中内接一等边三角形，它的一个顶点坐标是(0,1)，一个高在y 轴上，m n 、是互质的正整数.求m n +.6、 掷筛子四次，后三次的点数总不比前面的点数小的概率可以表示为/m n ，其中m n 、是互质的正整数.求m n +.7、 在A B C ∆中，21,22A B A C ==和20B C =.点D 和点E 分别在A B 与A C ,使得D E平行于B C 且D E 通过A B C ∆的内心.此时/D E m n =，其中m n 、是互质的正整数.求m n +.8、 如果一个十进制数N 在七进制下的结果是N 的2倍，就把它叫做7-10双倍数.例如51是7-10双倍数，因为它在七进制下的结果是102.求最大的7-10双倍数.9、 在A B C ∆中，13,15A B B C ==和17C A =.点D 、点E 和点F 分别在A B 、B C 和C A .设,A D p A B B E q B C =⋅=⋅和C F r C A =⋅，其中p q r 、、都是正数，且满足23p q r ++=，22225p q r++=.D E F ∆与A B C ∆的面积之比可以写成/m n 的形式，其中m n 、是互质的正整数.求m n +.10、设点集S 的坐标x y z 、、都是整数且满足02,03,04x y z ≤≤≤≤≤≤.从S 中随机抽取两个不同点，则二者的中点仍在S 中的概率为/m n ，其中m n 、是互质的正整数.求m n +.11、在5N ⨯点阵中，这些点从左到右、从上到下依次编号(如第一行的序号是从1到N ，第二行的序号是从1N +到2N ，等等).从第i 行中任意抽取一点i P ，这样我们选择了5个点1234,,,P P P P 和5P .设i P 对应数i x .现在从第一列开始重新对点进行从左到右的编序，这时设i P 对应数是i y .我们发现规律12213445,,,x y x y x y x y ====且53x y =.求N 的最小可能值.12、一个球内切于顶点分别是(6,0,0),(0,4,0),(0,0,2)A B C ===和(0,0,0)D =的四面体.球的半径是/m n ，其中m n 、是互质的正整数.求m n +.13、在某个特定的圆中，角度为d 的弧对应的弦长是22厘米，角度为2d 的弧对应的弦长比角度为3d 的弧对应的弦长大20厘米，其中120d <.角度为3d 的弧对应的弦长可以表示为m -+m n 、是互质的正整数.求m n +.14、邮递员要给街道上的19个房子送邮件.他发现没有两个相邻的房子在同一天都收到邮件，且连续多于两个的房子必定会有一个房子收到邮件.求有多少种可能邮递的方法? 15、数1,2,3,4,5,6,7和8随机的写在正八面体的8个面上，使得每个面上的数都不相同.则不相邻的两数共用一个边的概率是/m n ，其中m n 、是互质的正整数.求m n +. 答案 1、 答案 630用a 表示满足题目条件性质的十位数，b 表示个位数，故|10,|10a a b b a b ++，推出|,|10a b b a .前一个条件成立须满足b ka =，其中k 是某一正整数.后一条件暗含着1,2,5k k k ===.因此满足条件的两位数是11，22，33，...，90，12，24，36，48和15.求和结果为1145121015630⋅+⋅+=. 2、 答案 651设含有n 个元素的集合S 的平均值是x .故1131n x x n +=-+ 和2001271n x x n +=++即：1(1)13(1)n x n x n +=+-+ 和 2001(1)27(n x n x n +=+++.二者相减得200040(1)n =+，故49n =.因此651x =.3、 答案 500 应用二项式定理20012001200120012200120012000199920001999110()()2212001200012001...22220012001250...2xx xx xxxx xx=+-=--⋅⎛⎫⎛⎫=-+⋅-+ ⎪ ⎪⎝⎭⎝⎭=-⋅+根据根的和的公式得到220012505002001⋅⋅=.4、 答案291注意到角C 与角A T C 都是75︒，因此24A C A T ==。
2000 AIME I中文版
2000年美国数学邀请赛(1)1. 求最小的正整数n,使得不论怎样将10n 写成两个正整数的乘积,这两个正整数中至少有一个含有数码0.2. 令u 和v 是满足0<v<u 的整数,令A=(u,v),点B 与点A 关于直线y=x 对称,点C 与点B 关于y 轴对称,点D 与点C 关于x 轴对称,点E 与点D 关于y 轴对称,五边形ABCDE 的面积是451,求u+v.3. 在(ax+b)2000中,a 和b 是互质的正整数,其展开式中的x 2与x 3的系数相等,求a+b.4. 如图所示:一个长方形被分割成九个不重叠的正方形,已知长方形的长与宽是互质的正整数,求此长方形的周长.5. 有两个盒子,每个盒子中都是既有黑弹子,也有白弹子,两个盒子中的弹子的总数是25,从每个盒子中各取一粒弹子,取到两个黑弹子的概率是5027,取到两个白弹子的概率是nm,这里m 与n 是互质的正整数,求m+n.6. 有序整数组(x,y)有0<x<y<106,并且x 与y 的等差中项比等比中项大2,这样的数组有多少个?7. 设x,y,z 是正数,且满足xyz=1,x+z1=5,y+x1=29,若z+y1=nm ,这里m与n 是互质的正整数,求m+n.8. 一只外形是高为12英寸,底面半径是5英寸的直圆锥容器,圆锥的顶点在下,底面水平放置时,密闭在其中的液体深度是9英寸,当顶点在上,底面水平放置时,其中的液体的高度是m-n 3p,这里m,n,p是正整数,并且p 不能被任何质数的立方整除,求m+n+p.9. 方程组)x )(logz (log)zx (log1)z )(log y (log )yz 2(log 4)y )(logx (log)xy 2000(log101010101010101010=-=-=-有两组解,(x 1,y 1,z 1)和(x 2,y 2,z 2).求y 1+y 2.10. 数列x 1,x 2,x 3,⋯,x 100 有以下性质:对介于1到100(包括这两个数)的每个整数k,x k 比其余99个数的和少k.已知x 50=nm ,这里m 与n是互质的正整数,求m+n.11. 令S 是所有形如b a的数之和,这里a 和b 是互质的1000的约数,求不超过10S 的最大整数.12. 一种函数f 对任意实数x 都有f(x)=f(398-x)=f(2158-x)=f(3214-x),在f(0),f(1),f(2),⋯,f(999)有多少个不同的值.13. 在大草原的中央,有一辆消防车停在两条笔直的并且互相垂直的高速公路的交叉处.此消防车能以每小时50英哩的速度在高速公路上行驶,但只能以每小时14英哩的速度穿越草原.该消防车6分钟内能到达的点形成的区域的面积是n m平方英哩,这里m 与n 是互质的正整数,求m+n.14. 在∆ABC 中,已知∠B 与∠C 相等,点P 和点Q 分别在AC 和AB 上,使得AP=PQ=QB=BC.∠ACB 是∠APQ 的r 倍一样大,其中r 是正实数,求不超过1000r 的最大整数.15. 一叠卡片共2000张,每张卡片上标有1~2000中的某一个整数,不同的卡片上标着不同的数.这叠卡片不是按数的顺序排列的.将这叠卡片最上面的一张放到桌面上,并将紧接着的一张移到这叠卡片的最下面.然后将新的最上面的一张卡片放到桌面上,并且放在已经在那里的卡片的右边,再将紧接着的一张卡片移到这叠卡片的最下面.这样的过程(将最上面的卡片放到桌面上的卡片的右边,紧接着的卡片移到这叠卡片的最下面)不断重复,直到所有的卡片都放到桌面上.此后发现,桌面上的卡片从左到右是按递增的顺序1,2,3, ,2000放置的.在原来这叠卡片中,有多少张放在标有1999这个数的这张卡片的上面?答案:008 021 667 260 026997 005 052 025 173248 177 731 571 927。
2002 AIME I中文版
2002年美国数学邀请赛(1)1. 有许多州用三个数字再紧接三个字母作为标准的车牌形式.已知三个数字和三个字母的排列是等可能的,车牌中至少包含一个“回文”(三个数字或三个字母从左到右读与从右到左读是相同的)的概率是n m,这里m 和n 是互质的正整数,求m+n. 2. 如图所示:20个相同的圆在一个长方形里面排成三行,一个圆与另一个圆相切,与长方形的边相切,长方形较长的边的长度与较短的边的长度之比可以写成21(p -q),这里p 和q 是正整数,求p+q.3. 简今年25岁,迪克比简年长.n 年以后(这里n 是正整数),迪克的年龄和简的年龄都是两位数,简的年龄可由迪克交换其年龄数的两个数字来得到,令d 是迪克当前的年龄,有多少个可能的数对(d,n)?4. 考察由a k =k k 12+(k ≥1)定义的数列,已知满足m<n 的正整数m 和n有a m +a m+1+⋯+a n-1=291,求m+n.5. 设A 1,A 2,A 3,⋯,A 12是正十二边形的顶点,在这个十二边形所在的平面上,有多少个不同的正方形,它至少有两个顶点属于集合{A 1,A 2, A 3,⋯,A 12}?6. 方程组164log 225log 4y log x logy x 64225=-=+的解是(x 1,y 1),(x 2,y 2),求log 30(x 1y 1x 2y 2).7. 伯努利表达式对于非整数指数幂也是有效的.对任意的实数r 和x,y(|x|>|y|),(x+y)r =x r +rx r-1y 1+2)1r (r -x r-2y 2+!3)2r )(1r (r --x r-3y 3+⋯ 求7102002)110(+的十进制小数表示中小数点右面的前三位数字.8. 求最小的正整数k,使满足以下条件的数列不止一个.(1)a 1,a 2,a 3,⋯是不减的正整数数列;(2)当n>2时,a n =a n-1+a n-2;(3)a 9=k9. 哈罗德,塔妮雅和尤利塞斯为一段很长的围栏的柱子刷油漆.哈罗德从第一根柱子开始,并每次再刷第h 根柱子;塔妮雅从第二根柱子开始,并每次再刷第t 根柱子;尤利塞斯从第三根柱子开始,并每次再刷第u 根柱子,如果正整数数组(h,t,u)使得每根柱子恰好被刷了一次,则将正整数100h+10t+u 叫做“恰好可涂”数,求所有的“恰好可涂”数之和.10. 如图所示,∆ABC 是一个直角三角形,点D 在BC 上,AD 平分∠BAC,点E 和F 分别在AB 和AC 上,AE=3,AF=10,又已知EB=9,FC=27,求最接近四边形DCFG 面积的整数.11. ABCD 和BCFG 是棱长为12的正方体的两个面,一光束从A 点出发,在BCFG 面上的点P 处反射,P 到BG,BC 的距离分别是7和5,光线继续在正方体的各个面之间反射,它从A 点出发到下一次到达正方体的某个顶点的整个路径长为m n ,这里m 和n 是整数,n 不能被任何一个质数的平方整除,求m+n.12. 令F(z)=i z iz -+,对所有满足z ≠i 的复数z 并对所有正整数n,令z n =F(z n-1).已知z 0=1371+i , z 2002=a+bi,这里a 和 b 是实数,求a+b.13. 在∆ABC 中,中线AD 和中线CE 的长分别是18和27,AB=24,延长CE 与∆ABC 的外接圆相交于F,∆ABF 的面积是mn ,这里m 和n 是正整数,n 不能被任何质数的平方整除,求m+n.14. 由不同的正整数形成的集合S 有以下性质:对S 中的每个整数x,在S 中去掉x 所剩的数的算术平均数是一个整数,已知1属于S,2002是S 中的最大元素,S 中最多有多少个元素?15. 多面体ABCDEF 有六个面,面ABCD 是一个边长为12的正方形,面ABFG 是一个梯形,AB//GF,BF=AG=8,GF=6,面CDE 有CE=DE=14,其他面是ADEG,BCEF,EFG ,点E 到面ABCD 的距离是12,已知EG 2=p-qr ,这里p,q,r 是正整数,r 不能为任何质数的平方整除,求p+q+r.答案:059 154 025 840 183012 428 748 757 148230 275 063 030 163。
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108
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265
180
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275
840
728
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262
043
749
732
371
450
863
2017
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390
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581
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722
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307
507
341
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494
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2002
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059
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2018
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476018090Fra bibliotek440
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2016
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AIME历年真题答案
2000
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008
021
667
260